\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)

\(\Leftrightarrow x=\frac{2010}{10}=201\)

Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)

=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)

\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)

\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)

\(\Rightarrow x=\frac{2010}{10}=201\)

\(=\frac{1}{10}\)

=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

k mk nha

1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)

=> \(\frac{-1}{x+3}=\frac{1}{2010}\)

=> \(-x-3=2010\)

=> \(x=2007\)

\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)

\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)

\(x=\frac{1}{2011}:\frac{1}{2011}=1\)

Vậy x=1

\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)

\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)

\(\frac{1}{2011}.x=\frac{2}{4022}\)

\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)

Ai thấy đún thì ủng hộ mink nha !!!

Thanks you very much !!

Chúc các bạn luôn học giỏi !!!

a)\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)

\(A=\frac{1.2.3...19}{2.3.4...20}\)

\(A=\frac{1}{20}\)

đề bài thiếu bạn ơi

A=(2009/2010).(2008/2010). ... . (-2010/2010)

Còn lại mình chịu

mk chịu 

nhới k mk nha

\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)

\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(4-x+x=x\)

\(\Leftrightarrow x=4\)

lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!